Differential Calculus Questions and Answers

Answer: A

The tangent meets the x axis at \(\left(x_{0} - y_{0}/f'(x_{0}),\ 0\right)\).

The tangent at \((x_{0}, y_{0})\) has slope \(f'(x_{0})\):

\[ \begin{aligned} y - y_{0} &= f'(x_{0})(x - x_{0}) \\ \text{at } y = 0:\ x &= x_{0} - \frac{y_{0}}{f'(x_{0})} \end{aligned} \]

The condition \(f'(x) \neq 0\) is what guarantees the tangent is not horizontal and does cut the axis. This is the step used in Newton-Raphson iteration. The option with a plus sign has the sign of the correction wrong, and the two options with \(f'(x_{0})/y_{0}\) invert the ratio.

Answer: D

\(f(1)\) lies between 6 and (5 + e).

By the Mean Value Theorem there is some c with 0 < c < 1 such that

\[ \begin{aligned} f(1) - f(0) &= f'(c)(1 - 0) = e^{c} \\ 1 &\lt e^{c} \lt e \\ 6 &\lt f(1) \lt 5 + e \end{aligned} \]

Indeed \(f(x) = e^{x} + 4\) fits both conditions and gives \(f(1) = 4 + e \approx 6.72\), inside that interval. The other options forget to add \(f(0) = 5\), or add the wrong constant, so their bounds do not contain the value.

Answer: A

The function has a minimum at x = 1/2.

Set the derivative to zero and check the second derivative:

\[ \begin{aligned} f'(x) &= 2x - 1 = 0 \Rightarrow x = \tfrac{1}{2} \\ f''(x) &= 2 \gt 0 \end{aligned} \]

A positive second derivative means a minimum; the parabola opens upwards because the coefficient of \(x^{2}\) is positive, so it has no maximum at all. The minimum value is \(f(1/2) = -6.25\). The options at \(x = -1/2\) come from a sign slip in solving \(2x - 1 = 0\).

Answer: B

The interval is \(\left(\frac{\pi}{2}, \pi\right)\).

On \([0, 2\pi]\), \(\sin x\) decreases on \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\), where its derivative \(\cos x\) is negative, and \(\cos x\) decreases on \((0, \pi)\), where its derivative \(-\sin x\) is negative. Both decrease only on the overlap, \(\left(\frac{\pi}{2}, \pi\right)\), the second quadrant.

On \(\left(0, \frac{\pi}{2}\right)\) the sine is increasing; on \(\left(\pi, \frac{3\pi}{2}\right)\) the cosine is increasing; on \(\left(\frac{3\pi}{2}, 2\pi\right)\) both are increasing.

Answer: C

The function has minima at -2 and +2.

Write it as \(f(x) = (x^{2} - 4)^{2}\). As a square it is never negative, and it equals 0 exactly at \(x = \pm 2\), so both points are minima (the least possible value). Differentiating confirms the turning points:

\[ \begin{aligned} f'(x) &= 4x(x^{2} - 4) = 0 \\ x &= 0,\ \pm 2 \\ f''(x) &= 12x^{2} - 16 \end{aligned} \]

At \(x = \pm 2\), \(f'' = 32 \gt 0\), so both are minima; at \(x = 0\), \(f'' = -16 \lt 0\), a local maximum of value 16.

The function is even, \(f(-x) = f(x)\), so whatever happens at +2 happens at -2 too. That rules out the two options giving a minimum at one of them and a maximum at the other, and neither point can be a maximum since the function is zero there and positive nearby.

Answer: B

The limit is 1/2.

Substituting x = 0 gives 0/0, so rationalise the numerator by multiplying above and below by \(\sqrt{1+x} + 1\):

\[ \begin{aligned} \frac{\sqrt{1+x} - 1}{x} &= \frac{(1 + x) - 1}{x(\sqrt{1+x} + 1)} \\ &= \frac{1}{\sqrt{1+x} + 1} \to \frac{1}{2} \end{aligned} \]

The same value is the derivative of \(\sqrt{t}\) at t = 1, or the first term of the binomial expansion \(\sqrt{1+x} \approx 1 + x/2\). 0 and 1 come from mishandling the indeterminate form, and 2 from inverting the result.

Answer: A

The limit is \(-\frac{1}{3}\).

Expand numerator and denominator in powers of \(x\) near 0, using \(\cos x = 1 - x^{2}/2 + \cdots\) and \(\sin x = x - x^{3}/6 + \cdots\):

\[ \begin{aligned} x\cos x - \sin x &= x - \tfrac{x^{3}}{2} - x + \tfrac{x^{3}}{6} + \cdots \\ &= -\tfrac{x^{3}}{3} + \cdots \\ x^{2}\sin x &= x^{3} + \cdots \end{aligned} \]

The ratio therefore tends to \(-\frac{1}{3}\). The same result follows from L'Hopital's rule applied repeatedly to the 0/0 form.

\(\frac{1}{3}\) drops the sign: the numerator is negative for small positive \(x\), because \(x\cos x \lt \sin x\) there. 3 and \(-3\) invert the ratio.

Answer: C

The value is \(\frac{1}{4u}\).

Every term of \(u = x^{4} + y^{4} + 3x^{2}y^{2}\) has total degree 4, so \(u\) is a homogeneous function of degree 4. By Euler's theorem on homogeneous functions,

\[ \begin{aligned} &x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} \\ &= 4u \end{aligned} \]

Directly: \(x u_x = 4x^{4} + 6x^{2}y^{2}\) and \(y u_y = 4y^{4} + 6x^{2}y^{2}\), which add to \(4x^{4} + 4y^{4} + 12x^{2}y^{2} = 4u\). The question asks for the reciprocal, \((4u)^{-1} = \frac{1}{4u}\).

\(4u\) is the bracket itself, before the power \(-1\) is applied. \(2u\) uses the wrong degree, and 0 would hold only for a function of degree zero.

Answer: D

\(f\) is not differentiable at \(\{-1, 0, 1\}\).

The curves \(y = x\) and \(y = x^{3}\) cross where \(x^{3} = x\), at x = −1, 0 and 1. Between the crossings the larger one switches:

• \(x \lt -1\): \(x \gt x^{3}\), so f = x
• \(-1 \lt x \lt 0\): f = \(x^{3}\)
• \(0 \lt x \lt 1\): f = x
• \(x \gt 1\): f = \(x^{3}\)

At each switch the two pieces have different slopes (1 against \(3x^{2}\), which is 3 at x = ±1 and 0 at x = 0), so each is a corner. The options containing 2 or −2 name points where f is smooth, and \(\{0, 1\}\) misses the corner at −1.

Answer: D

The limit is -1.

Split the fraction:

\[ \begin{aligned} \frac{x^{3} - \sin x}{x} &= x^{2} - \frac{\sin x}{x} \\ &\to 0 - 1 = -1 \end{aligned} \]

using the standard limit \(\sin x / x \to 1\). L'Hospital's rule gives the same: \((3x^{2} - \cos x)/1 \to -1\). 1 drops the minus sign, 0 treats the whole numerator as negligible, and 3 comes from differentiating \(x^{3}\) but forgetting the sine term.